Question 1. Graphical Integration Challenge: $u$ -Substitution and Integration by Parts
Let $F (x) = \int_{- 1}^{x} f (t) d t$ and $G (x) = \int_{0}^{x} g (t) d t$ , where the graphs of $f (x)$ and $g (x)$ on $[- 2, 4]$ are below.
(a) Evaluate $\int_{0}^{2} [F (x) f (x) - G (x) g (x)] d x$ .
HINT: Consider a $u$ -substitution (possibly multiple substitutions). You may need to invoke Part I of the Fundamental Theorem of Calculus.
(b) Evaluate $\int_{0}^{1} f^{'} (G (x)) G (x) g (x)$ .
HINT: Consider a $u$ -substitution. You may also need to integrate by parts.

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$[f (G (x)) G (x)]_{0}^{1} - \int_{0}^{1} f (G (x)) g (x) d x$

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Step 1 ：Given that $F (x) = \int_{- 1}^{x} f (t) d t$ and $G (x) = \int_{0}^{x} g (t) d t$ , we are asked to evaluate the following integrals:

Step 2 ：(a) $\int_{0}^{2} [F (x) f (x) - G (x) g (x)] d x$

Step 3 ：Using the Fundamental Theorem of Calculus, we can simplify this integral to $F (2) f (2) - F (0) f (0) - G (2) g (2) + G (0) g (0)$

Step 4 ： $F (2) f (2) - F (0) f (0) - G (2) g (2) + G (0) g (0)$

Step 5 ：(b) $\int_{0}^{1} f^{'} (G (x)) G (x) g (x) d x$

Step 6 ：Using integration by parts and $u$ -substitution, we can simplify this integral to $[f (G (x)) G (x)]_{0}^{1} - \int_{0}^{1} f (G (x)) g (x) d x$

Step 7 ： $[f (G (x)) G (x)]_{0}^{1} - \int_{0}^{1} f (G (x)) g (x) d x$

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